The first ionization energy of a hydrogen atom is 1.31 kJ/mol. What is the wavelength of light, in nanometers, that is just sufficient to ionize a hydrogen atom? Refer to the values for constants.
The Correct Answer and Explanation is :
To determine the wavelength of light that is just sufficient to ionize a hydrogen atom, we use the relationship between energy and wavelength given by the equation:
[
E = \frac{hc}{\lambda}
]
where:
- (E) is the energy required to ionize a hydrogen atom (first ionization energy),
- (h) is Planck’s constant ((6.626 \times 10^{-34} \, \text{J·s})),
- (c) is the speed of light ((3.00 \times 10^8 \, \text{m/s})),
- (\lambda) is the wavelength in meters.
Step 1: Convert Energy to Joules
The given ionization energy is:
[
1.31 \times 10^3 \, \text{kJ/mol} = 1.31 \times 10^6 \, \text{J/mol}
]
Since this energy is per mole, we convert it to energy per atom using Avogadro’s number ((6.022 \times 10^{23}) atoms/mol):
[
E = \frac{1.31 \times 10^6 \text{ J/mol}}{6.022 \times 10^{23} \text{ atoms/mol}}
]
[
E = 2.176 \times 10^{-18} \text{ J/atom}
]
Step 2: Solve for Wavelength
Using (E = \frac{hc}{\lambda}), we rearrange for (\lambda):
[
\lambda = \frac{hc}{E}
]
Substituting the values:
[
\lambda = \frac{(6.626 \times 10^{-34} \text{ J·s}) (3.00 \times 10^8 \text{ m/s})}{2.176 \times 10^{-18} \text{ J}}
]
[
\lambda = 9.11 \times 10^{-8} \text{ m}
]
Converting meters to nanometers:
[
\lambda = 91.1 \text{ nm}
]
Final Answer:
The wavelength of light required to ionize a hydrogen atom is 91.1 nm.
This wavelength falls in the ultraviolet (UV) region, meaning hydrogen atoms require UV light for ionization.
Now, I will generate an image representing the hydrogen ionization process.
The image visually represents the ionization of a hydrogen atom, showing an ultraviolet (UV) photon striking the atom and ejecting its electron. This process illustrates how light with a wavelength of 91.1 nm provides sufficient energy to ionize the atom. Let me know if you need any modifications!
